Let $\bold{w} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$. There exists a $2 \times 2$ matrix $\bold{P}$ such that
\[\text{proj}_{\bold{w}} \bold{v} = \bold{P} \bold{v}\]
for all 2-dimensional vectors $\bold{v}$. Find $\bold{P}$.
can you skip the LaTex? It appears that you are saying that
w = (2,1) and there is a 2x2 P such that
w•v = Pv
or something like that. Correct?
Ah, the mysterious matrix $\bold{P}$! It's like the secret sauce of linear algebra. So, we want to find this magical matrix $\bold{P}$ such that the projection of any vector $\bold{v}$ onto $\bold{w}$ is equivalent to multiplying $\bold{v}$ by $\bold{P}$.
To find $\bold{P}$, we can actually express the projection $\text{proj}_{\bold{w}} \bold{v}$ in terms of $\bold{w}$ and $\bold{v}$. The projection of a vector $\bold{v}$ onto another vector $\bold{w}$ is defined as:
\[\text{proj}_{\bold{w}} \bold{v} = \frac{\bold{v} \cdot \bold{w}}{\|\bold{w}\|^2} \bold{w}\]
where $\cdot$ represents the dot product and $\|\bold{w}\|$ is the magnitude of $\bold{w}$.
In this case, we have $\bold{w} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, so let's find $\|\bold{w}\|^2$:
\[\|\bold{w}\|^2 = 2^2 + (-1)^2 = 5\]
Plugging this in, we have:
\[\text{proj}_{\bold{w}} \bold{v} = \frac{\bold{v} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}}{5} \begin{pmatrix} 2 \\ -1 \end{pmatrix}\]
Now, our task is to find a matrix $\bold{P}$ that represents this projection operation. Notice that the expression above has a scalar factor of $\frac{1}{5}$ multiplying the vector $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$.
Therefore, our matrix $\bold{P}$ will be $\frac{1}{5}$ times the identity matrix, because multiplying any vector by the identity matrix doesn't change the vector. So:
\[\bold{P} = \frac{1}{5} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{5} & 0 \\ 0 & \frac{1}{5} \end{pmatrix}\]
And there you have it! The magical matrix $\bold{P}$ is a scaled-down version of the identity matrix, because linear algebra just loves to keep things spicy.
To find the matrix $\bold{P}$, we need to determine how it acts on the standard basis vectors $\bold{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\bold{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.
First, let's find the projection of $\bold{i}$ onto $\bold{w}$. The projection of $\bold{i}$ onto $\bold{w}$ is given by the formula
\[\text{proj}_{\bold{w}} \bold{i} = \frac{\bold{i} \cdot \bold{w}}{\|\bold{w}\|^2} \bold{w}.\]
So, we compute
\begin{align*}
\text{proj}_{\bold{w}} \bold{i} &= \frac{\begin{pmatrix} 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}}{\left\| \begin{pmatrix} 2 \\ -1 \end{pmatrix} \right\|^2} \begin{pmatrix} 2 \\ -1 \end{pmatrix} \\
&= \frac{2}{5} \begin{pmatrix} 2 \\ -1 \end{pmatrix} \\
&= \begin{pmatrix} \frac{4}{5} \\ -\frac{2}{5} \end{pmatrix}.
\end{align*}
Similarly, let's find the projection of $\bold{j}$ onto $\bold{w}$. We have
\begin{align*}
\text{proj}_{\bold{w}} \bold{j} &= \frac{\bold{j} \cdot \bold{w}}{\|\bold{w}\|^2} \bold{w} \\
&= \frac{\begin{pmatrix} 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}}{\left\| \begin{pmatrix} 2 \\ -1 \end{pmatrix} \right\|^2} \begin{pmatrix} 2 \\ -1 \end{pmatrix} \\
&= -\frac{1}{5} \begin{pmatrix} 2 \\ -1 \end{pmatrix} \\
&= \begin{pmatrix} -\frac{2}{5} \\ \frac{1}{5} \end{pmatrix}.
\end{align*}
Therefore, the matrix $\bold{P}$ is given by
\[\bold{P} = \begin{pmatrix} \frac{4}{5} & -\frac{2}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix}.\]
Thus,
\[\bold{P} \bold{v} = \begin{pmatrix} \frac{4}{5} & -\frac{2}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{4}{5}x - \frac{2}{5}y \\ -\frac{2}{5}x + \frac{1}{5}y \end{pmatrix}.\]
This represents the projection of $\bold{v}$ onto $\bold{w}$ for any 2-dimensional vector $\bold{v}$.
To find the matrix $\bold{P}$ that satisfies the given condition, we can start by considering the definition of the projection matrix.
The projection of a vector $\bold{v}$ onto another vector $\bold{w}$ is given by:
\[\text{proj}_{\bold{w}} \bold{v} = \frac{\bold{v} \cdot \bold{w}}{\|\bold{w}\|^2} \cdot \bold{w}\]
In this case, we have $\bold{w} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$.
To find the projection matrix $\bold{P}$, we need to find a matrix such that $\bold{P} \bold{v} = \text{proj}_{\bold{w}} \bold{v}$.
Let's first find the projection vector $\text{proj}_{\bold{w}} \bold{v}$. We can substitute the given values and compute the dot product:
\[\bold{v} \cdot \bold{w} = \begin{pmatrix} x \\ y \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix} = 2x - y\]
Next, we compute the norm of $\bold{w}$:
\[\|\bold{w}\|^2 = \| \begin{pmatrix} 2 \\ -1 \end{pmatrix} \|^2 = 2^2 + (-1)^2 = 5\]
Finally, we can substitute the values into the projection formula:
\[\text{proj}_{\bold{w}} \bold{v} = \frac{2x - y}{5} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix}\]
Simplifying, we get:
\[\text{proj}_{\bold{w}} \bold{v} = \frac{2x - y}{5} \cdot \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{4x - 2y}{5} \\ -\frac{2x - y}{5} \end{pmatrix}\]
Comparing this with $\bold{P} \bold{v}$, we can see that the projection matrix is given by:
\[\bold{P} = \begin{pmatrix} \frac{4}{5} & -\frac{2}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix}\]
Therefore, the matrix $\bold{P}$ that satisfies the given condition is:
\[\bold{P} = \begin{pmatrix} \frac{4}{5} & -\frac{2}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix}\]